Question: The differentiable functions $x$ and $y$ are related by the following equation: $y^2=7x+1$ Also, $\dfrac{dx}{dt}=-3$. Find $\dfrac{dy}{dt}$ when $y=6$.
Let's start by differentiating the equation $y^2=7x+1$ with respect to $t$. $\begin{aligned} y^2&=7x+1 \\\\ 2y\cdot\dfrac{dy}{dt}&=7\cdot\dfrac{dx}{dt} \end{aligned}$ We are given that $\dfrac{dx}{dt}=-3$, and we want to find $\dfrac{dy}{dt}$ when $y=6$. Let's plug ${y=6}$ and ${\dfrac{dx}{dt}=-3}$ into the equation we obtained: $\begin{aligned} 2{y}\cdot\dfrac{dy}{dt}&=7\cdot{\dfrac{dx}{dt}} \\\\ 2({6})\cdot\dfrac{dy}{dt}&=7({-3}) \\\\ 12\cdot\dfrac{dy}{dt}&=-21 \\\\ \dfrac{dy}{dt}&=-1.75 \end{aligned}$ In conclusion, when $y=6$, the value of $\dfrac{dy}{dt}$ is $-1.75$.